How to find the Maclaurin series and the radius of convergence for #f(x)=1/(1+x)^2#?

#1/(1-x)=sum_(k=0)^oox^k=1+x+x^2+x^3+...#

Replace #x# with #(-x)#:

#1/(1+x)=1/(1-(-x))=sum_(k=0)^oo(-x)^k=sum_(k=0)^oo(-1)^kx^k#

Note that the derivative of #1/(1+x)=(1+x)^-1# is #-(1+x)^-2=(-1)/(1+x)^2#. We should then differentiate the series:

#(-1)/(1+x)^2=d/dxsum_(k=0)^oo(-1)^kx^k=sum_(n=0)^oo(-1)^kd/dxx^k=sum_(k=0)^oo(-1)^k(kx^(k-1))#

We can move the index of the sum to start at #k=1#, since the term #k=0# results in an addend of just #0#.

We also multiply by a factor of #-1# to find the desired result:

#1/(1+x)^2=-sum_(k=1)^oo(-1)^kkx^(k-1)=sum_(k=1)^oo(-1)^(k+1)kx^(k-1)#

#=1-2x+3x^2-4x^3+...#

Which has radius of convergence #R=1#, since the original substitution of #-x# for #x# would not affect the radius, and neither would the differentiation or multiplication by a factor of #-1#.

To find the Maclaurin series for \( f(x) = \frac{1}{{(1+x)^2}} \), we can first express \( f(x) \) in terms of its derivative, then find the Maclaurin series representation using the known Maclaurin series expansions. First, let's find the first few derivatives of \( f(x) \): \( f(x) = \frac{1}{{(1+x)^2}} \) \( f'(x) = \frac{-2}{{(1+x)^3}} \) \( f''(x) = \frac{6}{{(1+x)^4}} \) \( f'''(x) = \frac{-24}{{(1+x)^5}} \) Now, observe the pattern in the derivatives. We see that each derivative involves a factorial term in the denominator and alternates in sign. Specifically, \( f^{(n)}(0) = (-1)^n \cdot n! \). The Maclaurin series expansion for a function \( f(x) \) is given by: \( f(x) = f(0) + \frac{{f'(0)}}{{1!}}x + \frac{{f''(0)}}{{2!}}x^2 + \frac{{f'''(0)}}{{3!}}x^3 + \ldots \) For \( f(x) = \frac{1}{{(1+x)^2}} \), we have: \( f(0) = 1 \) \( f'(0) = -2 \) \( f''(0) = 6 \) \( f'''(0) = -24 \) Substitute these values into the Maclaurin series formula: \( f(x) = 1 - 2x + 6x^2 - 24x^3 + \ldots \) This is the Maclaurin series for \( f(x) = \frac{1}{{(1+x)^2}} \). To find the radius of convergence, we can use the Ratio Test. The Ratio Test states that if \( \lim_{{n \to \infty}} \left| \frac{{a_{n+1}}}{{a_n}} \right| = L \), then the series converges if \( L < 1 \) and diverges if \( L > 1 \). In our case, the Maclaurin series for \( f(x) = \frac{1}{{(1+x)^2}} \) is: \( f(x) = 1 - 2x + 6x^2 - 24x^3 + \ldots \) The ratio of consecutive terms is: \( \left| \frac{{a_{n+1}}}{{a_n}} \right| = \frac{{|(-1)^{n+1} (n+1)!|}}{{|(-1)^n n!|}} = \frac{{n+1}}{{n}} \) As \( n \to \infty \), this ratio approaches 1. Therefore, the radius of convergence is 1.